General Chemistry Question

[DrTacoElf]

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What volume of 1M HCl was added if 20.0ml of 1M NaOH is titrated to produce a pH of 2?


a)10.2 ml [Wrong]
b)20.2 ml [?]
c)30.4 ml [?]
d)35.5 ml [?]
e)none of these [?]


Ok here is what i can deduce. First to reach neutralization 20.0ml 1M HCl must be added to 20.0ml NaOH. This yields 40.0ml of solution and at a .5M concentration. Next final [H+] has to be .01M (pH=2). So

MV = MV
(1)(x) = (.01)(40+x)
Solved X = .4 (additional Hcl to be added)
Final volume of HCl would then be 20+.4 = 20.4?

So none of these 😕

 

[captain hazel]

yah. sounds good to me. I did it a tiny bit differently & got the same answer. (2=-log[(x-.02)mol/(x+.02)L] gives x=.0204 L).

I guess if you go backwards from your answer, you'd get .0004 mol of unneutralized H+ in .0404 L. And -log(4/400) = 2.

Why, was the credited answer something else? cause you seem right to me.

 

[DrTacoElf]

No i just hate none of these b/c you never know if you did it right...

 

[JavadiCavity]

You need to end up with .01 mol/L of H+ in solution in order to get a pH of 2. Since you .02 mols are used up in neutralizing the base, you would at least need to start with .03 mols/L of the acid. Without doing the calculations, I'd surmise that if you add 30 mL of 1M HCL, you'd be adding about .03 moles of H+ to the base. After neutralization, you'd have .01 mols in 50 mL of solution. Thus, 30 mL of 1M HCL gives you an answer pretty close to (C).

Sorry if this is wrong. Just a quick guess is all!

 

[blankguy]

Strong base with strong acid in equal amounts is PH of 7 so that eliminates a and b.

If we use the amount in c( in liters)

1M*.0304 liters=.0304mol HCl

1M*.020 liters=.020mol NaOH

.0304-.020=.0104mol HCl left after all the NaOH is neutralized

the rxn we are looking at is HCl with the Pka(look it up in a table)

The concentration for the left over HCl is .0104/(.020+.0304)
Setup the equilibrium pKa equation and find the concentration for the H+
or H3O+ concentration. Take its log to get its PH

 

[DrTacoElf]

Strong base with strong acid in equal amounts is PH of 7 so that eliminates a and b.

If we use the amount in c( in liters)

1M*.0304 liters=.0304mol HCl

1M*.020 liters=.020mol NaOH

.0304-.020=.0104mol HCl left after all the NaOH is neutralized

the rxn we are looking at is HCl with the Pka(look it up in a table)

The concentration for the left over HCl is .0104/(.020+.0304)
Setup the equilibrium pKa equation and find the concentration for the H+
or H3O+ concentration. Take its log to get its PH

good method but you won't have the Pka on the DAT.

 

[blankguy]

HCl is a very strong acid. It would safe to assume it's PKa is around 1, if I remember correctly the table of acids and bases in the chem textbook that's where HCl falls under. I take the DAT is going to be dealing with strong acids and strong bases, otherwise it becomes a bit hairy?